Q.1- One mole of an ideal gas undergoes a cyclic change ABCD where the (P, V) coordinate are (5,1), B(5,3), C(2,3) and D(2,1) where P is in atmosphere and V is in litre. Calculate work done along AB, BC, CD and DA and also net work done in the process. Given 1 atmosphere = 1.01 x 105 N/m2

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward

Answer(a)   the matererials that are in contact, the two materials and the nature of their surfaces/// (b)/// the force pushing the two surfaces together, pushing the surfaces together causes the more of the asperities to come together and increases the surface area in contact with each other

So are u adding some additional catalyst